848 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th15-4.pm5

Fig. 15.54

In this case F1–2 = F2–3 = 1 ; and

A

A

r

r

1

2

1

2

5

10

== = 0.5

A

A

r

r

2

3

2

3

10

15

=== 0.67

∴

2 1000

100 100

1005

005

1 1005

005

05

1

4

2

4

πrLFHG JIK −FHGT IKJ

L

N

M

M

O

Q

P

P

F −

HG

I

KJ

++F −

HG

I

KJ

. ×
.

.

.

.

2

100

300

100

1005

005

1 1005

005

0

2 2

44

πrLFHGT IKJ −FHG IKJ

L

N

M

M

O

Q

P

P

F −

HG

I

KJ++

F −

HG

I

KJ×

.

.

.

.

.67

or

0 05 10000

29.4

081

32.73

.( −^44 ) )

= −

xx.1(

or (1000 – x^4 ) =

29.5 0.1

32.73 0.05

×

× (x

(^4) – 81) = 1.8(x (^4) – 81)
or 2.8x^4 = 10000 – 145.8 = 9854.2
or x =
9854.2
2.8
F
HG
I
KJ
1/ 4
= 7.7
or
T 2
100 = 7.7 or T^2 = 770 K
∴ Heat flow per m^2 area of cylinder 1,
Q 12 =
AT T
A
A
114 24
1
1
2
2
1
2
1 1 1
σ
ε
ε
ε
ε
()−
F −
HG
I
KJ
++−
F
HG
I
KJ

15.67^1000
100
770
100
10.05
0.05
1 10.05
0.05
0.5
44
× GFH IKJ −FHG IKJ
L
N
M
M
O
Q
P
P
F −
HG
I
KJ
++F −
HG
I
KJ
×
= 5.67 (10000 3515.3)
29.5
×− = 1246.4 W. (Ans.)
Example 15.30. Two concentric spheres 210 mm and 300 mm diameters with the space between
them evacuated are to be used to store liquid air (– 153°C) in a room at 27°C. The surfaces of the
spheres are flushed with aluminium (ε = 0.03) and latent heat of vaporization of liquid air is
209.35 kJ/kg. Calculate the rate of evaporation of liquid air. (M.U.)
Solution. Given : r 1 =
210
2
= 105 mm = 0.105 m ; r 2 =
300
2
= 150 mm = 0.15 m ;
T 1 = – 153 + 273 = 120 K ; T 2 = 27 + 273 = 300 K ; ε 1 = ε 2 = 0.03, hfg = 209.35 kJ/kg.